Science:Math Exam Resources/Courses/MATH152/April 2012/Question 08 (c)/Solution 1

The complex solution ${\vec {\mathbf {z} }}(t)$ that we found in (a) and (b) is

${\begin{aligned}{\vec {\mathbf {z} }}(t)&=e^{-t}e^{2ti}{\begin{pmatrix}1\\-i\end{pmatrix}}\\&=e^{-t}(\cos 2t+i\sin 2t){\begin{pmatrix}1\\-i\end{pmatrix}}\\&=e^{-t}{\begin{pmatrix}\cos 2t\\\sin 2t\end{pmatrix}}+ie^{-t}{\begin{pmatrix}\sin 2t\\-\cos 2t\end{pmatrix}}\end{aligned}}$

From this we can read off the real-valued form as

${\vec {\mathbf {x} }}(t)=C_{1}e^{-t}{\begin{pmatrix}\cos 2t\\\sin 2t\end{pmatrix}}+C_{2}e^{-t}{\begin{pmatrix}\sin 2t\\-\cos 2t\end{pmatrix}}$